Differentiate $\sqrt{\frac{1 +\sin x}{1 -\sin x}}$

Question

I have tried a lot of ways to solve this question but I am unable to get the answer as same as my textbook.

The text book answer is as follow: $$\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)$$

The steps which I took is as follows:

$$\sqrt{\frac{1+ \sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}}$$

Then Secondly

$$\sqrt{\frac{\left(1+\sin x\right)^2}{1-\sin^2 x}}$$

Then I got

$$\dfrac{1+\sin x}{\cos x}$$

When I differentiated this I got the following

$$\frac{\cos ^2\left(x\right)+\sin \left(x\right)\left(1+\sin \left(x\right)\right)}{\cos ^2\left(x\right)}$$

Can Anyone tell me what I am doing wrong?

I also know that $$\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)=\frac{2}{\left(\cos \frac{x}{2}-\sin\frac{x}{2}\right)^2}$$

Thank you for the help!

Answer 1

Why make things complicated if there is a easy way? By the half-angle formulae we obtain

$$\sqrt{\frac{1-\sin(x)}{1+\sin(x)}}=\sqrt{\frac{1-\cos\left(x+\frac\pi2\right)}{1+\cos\left(x+\frac\pi2\right)}}=\tan\left(\frac x2-\frac\pi4\right)$$

And I suppose you can differentiate the tangent function ;)

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As pointed out by Simply Beautiful Art and mathcounterexamples.net by using the half-angle formula we ran into serious issus concerning the sign.

Answer 2

Logarithmic differentiation makes things easier $$y=\sqrt{\dfrac{1 +\sin (x)}{1 -\sin (x)}}\implies \log(y)=\frac 12 \left(\log(1+\sin(x)) -\log(1-\sin(x))\right)$$ $$\frac {y'}{y}=\frac 12 \left(\frac{\cos(x)}{1+\sin(x) }+\frac{\cos(x)}{1-\sin(x) }\right)$$ Simplify as much as you can and, when finished, use $$y'=y\times \frac {y'}{y}$$

Answer 3

Alternatively, using the product rule: $$\begin{align}\left(\sqrt{\dfrac{1 +\sin x}{1 -\sin x}}\right)' &=(\sqrt{1+\sin x})'\cdot (1-\sin x)^{-1/2}+\sqrt{1+\sin x}\cdot ((1-\sin x)^{-1/2})'=\\ &=\frac{\cos x}{2\sqrt{1+\sin x}}\cdot \frac1{\sqrt{1-\sin x}}+\sqrt{1+\sin x}\cdot \frac{\cos x}{2(1-\sin x)\sqrt{1-\sin x}}=\\ &=\frac{\cos x}{2\sqrt{\cos ^2x}}+\frac{\cos x\sqrt{(1+\sin x)^2}}{2(1-\sin x)\sqrt{1-\sin ^2x}}=\\ &=\frac12+\frac{1+\sin x}{2(1-\sin x)}=\\ &=\frac1{1-\sin x}=\cdots =\\ &=\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)\end{align}$$ Can you show the equality of the last two expressions using what you stated you know?

Answer (see the hidden area):

$$\frac1{1-\sin x}=\frac{1}{\sin^2x+\cos^2x-2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{1}{(\sin \frac x2-\cos \frac x2)^2}=\\=\frac{1}{2(\frac{1}{\sqrt{2}}\sin \frac x2-\frac{1}{\sqrt{2}}\cos \frac x2)^2}=\frac{1}{2\cos^2(\frac{\pi}{4}+\frac x2)}=\frac12\sec^2(\frac{\pi}{4}+\frac x2).$$

Answer 4

What if $1\pm\sin x=0?$

Otherwise $$\sqrt{\dfrac{1+\sin x}{1-\sin x}}=\sqrt{\left(\dfrac{1+\tan\dfrac x2}{1-\tan\dfrac x2}\right)^2}=\left|\tan\left(\dfrac\pi4+\dfrac x2\right)\right|$$ using
https://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml

Now $\tan\left(\dfrac\pi4+\dfrac x2\right)$ will be $>0$ if $1-\tan^2\dfrac x2>0\iff-1<\tan\dfrac x2<1$

Answer 5

The textbook answer may not be quite right.
For x in range $[0, 2\pi],\text{ only } [0, {\pi \over 2}], [{3\pi \over 2},2\pi]$ work.

This is a revised derivative, to handle the full range.

$$\left(\sqrt{{1+\sin(x) \over 1-\sin(x)}} \right)' = {sign(\cos(x))\over 2} \sec^2({\pi \over 4} + {x \over 2})$$

Of course, chain rule result works too, but a bit messy.
After some simplification, this is what I have.
Note: derivative have the sign of $\cos(x)$, as expected. $$\left(\sqrt{{1+\sin(x) \over 1-\sin(x)}} \right)' = {\cos(x) \over (1-\sin(x))^2 \sqrt{{1+sin(x) \over 1-sin(x)}}}$$